218 Mr. Babbage’s essay towards 
I have also made some attempts at discovering particular 
solutions of the two following equations, and have met with 
no success. 
4 , (x,y) = 4 ^’ 2 (x,y) and ^ 2 > 2 (x,y) = 1 (x,y) 
Should, however, any particular case be found, their general 
solutions flow immediately from the method just explained. 
With regard to the equation of the Problem ^ 2 > l (x,y) = 
'P J>2 (x,y) i I have little expectation of finding any particular 
case, as I think the following reasoning, though perhaps not 
quite so satisfactory as might be wished, will show the impos- 
sibility of it. First, let us suppose, that (x,y) is a symme- 
trical function of x andy, let it be % (x, y) then our equation 
becomes 
x { x C x >y),y] = %{x,x C x ~y ') } — % { z (*. y ), x } 
Comparing the first of these expressions with the third, 
we may observe that in the first, wherever % (x, y) occurs, the 
same quantity % (x, y) also occurs in the third, consequently 
in this respect, the first and third are identical : but wherever 
y occurs in the first, x occurs similarly in the third, therefore 
in this respect they cannot be identical, unless y is equal to x. 
From this it appears, that the equation in question cannot be 
solved by any symmetrical function. Again, the given equa- 
tion 4' 2> '(x, y) = 4' x » 2 {x,y) contains x andy in the same man- 
ner, and no reason can be assigned why in the solution x 
should be contained differently from y: this may, perhaps, 
be made more clear, thus. Let / (x, y) be the quantity to 
which each side of the given equation is equal, then 
<!/’>■ (x,y) =J{x,y) = X'*( x ,y) 
Now since ^ (4 { x > y)>lf) —S i x > !/) an ^ also 4 i x > 4* { x > y))= 
